Let $f(x) = \dfrac{2\sqrt{1+x^2}-5\sqrt{1-x^2}}{5\sqrt{1+x^2}+2\sqrt{1-x^2}}$.
Hence, find $\frac{dy}{dz}$ when $y=\cot^{-1}(f(x))$ with respect to $z=\cos^{-1}{\sqrt{1-x^4}}$.
To get this into a simpler form, I tried the substitution $x=\sqrt{\sin\theta}$ which reduces $z$ to $\theta$. But this substitution has no effect on $y$. I haven't been able to reduce $y$ by trying to get it into the form $\dfrac{\tan A+\tan B}{1-\tan A·\tan B}$. I tried $1-\sin\theta=|(\sin\frac{\theta}{2} - \cos\frac{\theta}{2})^2|$, but which sign of the modulus should am I supposed to go with? How to go about this?
Divide the numerator & the denominator by $5\sqrt{1+x^2}$
and choose $\tan A=\dfrac25,\tan B=\sqrt{\dfrac{1-x^2}{1+x^2}}\implies x^2=\cos2B$
$\implies f(x)=\tan u=\tan(A-B)\implies u=A-B$
$\implies \arctan\{f(x)\}=A-\dfrac12\arccos(x^2)$
$f'(x)=\dfrac x{\sqrt{1-x^4}\cdot\{1+f^2(x)\}}$